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Assume that $\Gamma$ is a group with neutral element $e$. We associate to $\Gamma$ the following groupoid $G$:

$G=\Gamma \times \Gamma,\;\;\;G^{(0)}=\Gamma \times \{0\},\;\;s(a,b)=(a,e),\;\;\; r(a,b)=(ba, e)$

If $\phi:\Gamma_{1}\to \Gamma_{2}$ is a group isomorphism, then $\tilde{\phi}:G_{1} \to G_{2}$ with $\tilde{\phi}(a,b)=(\phi(a), \phi(b))$ is a groupoid isomorphism. So isomorphism groups give us isomorphic groupoids. Now we ask the converse:

Are there two non isomorphic groups $\Gamma_{1}, \Gamma_{2}$ such that the corresponding groupoids $G_{1}, G_{2}$ are isomorphic.

Note that a groupoid isomorphism between $G_{1}, G_{2}$ does not necessarily come from a group isomorphism between $\Gamma_{1}, \Gamma_{2}$, as constructed above. An easy example can be provided by $\Gamma_{1}=\Gamma_{2}=\mathbb{Z}/2\mathbb{Z}$.

This situation is a motivation for the above question.

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    $\begingroup$ Your groupoid is the action groupoid of $G$ acting on itself by left multiplication, and accordingly it is contractible: that is, it's equivalent to the terminal groupoid $1$. (Do you really want to ask for an isomorphism rather than an equivalence of groupoids? It's not a particularly natural condition.) $\endgroup$
    – Qiaochu Yuan
    Commented Jan 22, 2016 at 19:04
  • $\begingroup$ @QiaochuYuan yes I mean "isomorphism between groupoids". $\endgroup$
    – Ali Taghavi
    Commented Jan 22, 2016 at 19:06

1 Answer 1

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Look at the group $Aut(x)$ where $x$ is an object of the groupoid. You find that this group is trivial. Therefore your groupoid is THE groupoid with trivial automorphism groups and its isomorphy class only depends on the cardinality of the set of objects. Hence any two groups of the same cardinality will give isomorphic groupoids

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  • $\begingroup$ Than you for your answer.I do not understand why such groupoid has no non identity automorphism. For example put $\Gamma=\mathbb{Z}/2\mathbb{Z}$ then $G=\Gamma \times \Gamma $ has non identity automorphism $\phi(0,0)=(1,0),\;\phi(1,0)=(0,0), \phi(1,1)=(0,1),\; \phi(0.1)=(1,1)$. Am I missing some thing? $\endgroup$
    – Ali Taghavi
    Commented Jan 22, 2016 at 19:53
  • $\begingroup$ I don't understand what you mean by "identity automorphism".The groupoid has many automorphisms, indeed, every permutation of the object set gives one. The objects, however, have no automorphisms. $\endgroup$
    – user1688
    Commented Jan 22, 2016 at 21:11
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    $\begingroup$ What Anton means is the set of automorphisms of a single object in your groupoid. This will always be a group, for trivial reasons. In a given connected component all objects have isomorphic automorphism groups (again for trivial reasons). Furthermore, the $hom(x,y)$ sets for distinct $x,y$ in a connected components are $Aut(x)$-torsors, i.e. copies of $Aut(x)$ without a distinct identity element. Since the action groupoid of a group acting on itself is connected and has trivial $Aut(x)$, the groupoid is completely determined up to isomorphism by its cardinality. $\endgroup$
    – Georg Lehner
    Commented Jan 22, 2016 at 22:03
  • $\begingroup$ @GeorgLehner thank you for your comment. It help me to understand the details. $\endgroup$
    – Ali Taghavi
    Commented Jan 23, 2016 at 8:26
  • $\begingroup$ @Anton thanks again for your answer. $\endgroup$
    – Ali Taghavi
    Commented Jan 23, 2016 at 8:26

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